Question

as a system increases in volume, it absorbs 58.0 J of energy in the form of heat from the surroundings. the piston is working against a pressure of 0.556 atm. the final volume of the system is 56.3 L. what was the initial volume of the system if the internal energy of the system decreased by 101.7 J?

Answer 1

According to the 1st law of thermodynamics:

U = Q - W, where U is the change in internal energy, Q is the heat exchange and W is the work done.

The sign conventions for Q and W are :

Q is positive if heat is added to the system and negative if heat is removed.

W is positive if work is done by the system and negative if work is done on the system.

In our problem, since internal energy decreases, U = -101.7 J

Since heat is added/absorbed to the system, Q = 58 J

W = pdV, where p is the pressure the piston is working against and dV is the change in volume

dV = V_final - V_initial = (56.3 - V) L.

Now, 1 L-atm = 101.325 J,

Thus, pdV = 0.556 (56.3 - V) L-atm = 101.325 x 0.556 (56.3 - V) J

Using the 1st law written at start,

-101.7 J = 58 J - 56.34( 56.3 -V) J

56.3 -V = (101.7 + 58)/56.34

V= 56.3 L - 2.83 L

V = 53.47 L

Thus, the initial volume of the system is 53.47 L